What is the integral of (e^{1/(x^2)})/(x^3) ?

The integral of (e^{1/(x^2)})/(x^3) is -1/2 e^{1/(x^2)}+C

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integral of (e^{1/(x^2)})/(x^3)

\int \frac{e ^{\frac{1}{x ^{2}}}}{x ^{3}} d x

- \frac{1}{2} e^{\frac{1}{x ^{2}}} + C

Solution steps

\int \frac{e ^{\frac{1}{x ^{2}}}}{x ^{3}} d x

Apply u-substitution

: \int \frac{e ^{\frac{1}{u}}}{2 u ^{2}} d u

= \int \frac{e ^{\frac{1}{u}}}{2 u ^{2}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{e ^{\frac{1}{u}}}{u ^{2}} d u

Apply u-substitution

: \int - e^{v} d v

= \frac{1}{2} \cdot \int - e^{v} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} (- \int e^{v} d v)

Use the common integral:

\int e^{v} d v = e^{v}

= \frac{1}{2} (- e^{v})

Substitute back

= \frac{1}{2} (- e^{\frac{1}{x ^{2}}})

Simplify

= - \frac{1}{2} e^{\frac{1}{x ^{2}}}

Add a constant to the solution

= - \frac{1}{2} e^{\frac{1}{x ^{2}}} + C

ln (3 x + 7)

\frac{d}{d x} (∣ - 3 x ∣)

y^{^{'}} + 9 y = t + e^{- 8 t}

\frac{\partial}{\partial t} (e^{h (t)} sin (m x) cos (n y))

(sin (x y) + x y cos (x y)) d x + x^{2} cos (x y) d y = 0

What is the integral of (e^{1/(x^2)})/(x^3) ?
The integral of (e^{1/(x^2)})/(x^3) is -1/2 e^{1/(x^2)}+C